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Class 10 Mathematics Quadratic Equations Extra Questions

Chapter 4: Quadratic Equations

Topic–1: Quadratic Equations

Very Short Answer Type Questions: [1 Mark]

1. If 8 is the root of the equation x2 – 10x + k = 0, then find the value of k.

Sol. (8)2 – 10 × 8 + k = 0

⇒ 64 – 80 = –k ⇒ k = 16

2. Is x = 3 a root of the equation 2x2 – 5x – 3 = 0?

Sol. 2(3)2 – 5(3) – 3 = 0

⇒ 18 – 15 – 3 = 0

So, x = 3 is the root of 2x2 – 5x – 3 = 0.

3. (x + 2)3 = x(x2 – 1) is a quadratic equation. Is it true?

Sol. (x + 2)3 = x(x2 – 1)

⇒ x3 + 12x + 6x2 + 8 = x3 – x

⇒ 6x2 + 13x + 8 = 0

So, it is a quadratic equation.

4. Is x = 1 a root of the equation 3x2 – 2x – 1 = 0?

Sol. 3(1)2 – 2 × 1 – 1 = 3 – 2 – 1 = 3 – 3 = 0

So, x = 1 is a root of 3x2 – 2x – 1 = 0

5. For what value of k, x = 2 is a solution of kx2 + 2x – 3 = 0?

Sol. k(2)2 + 2 × 2 – 3 = 0

⇒ 4k + 4 – 3 = 0

 k = − ¼

 6. If x = 2 is a root of the equation 3×2 – 2kx + 5 = 0, then find k.

Sol. 3 × (2)2 – 2 × 2 × k + 5 = 0

⇒ 12 – 4k + 5 = 0 ⇒ k = 17/4

Short Answer Type Questions:

  1. Which of the following among the given expressions is a polynomial?

 i) 1 + √x + x        (ii) 1 + x2        (iii) x + 1/x

 Sol. (i) The power of x in √x, that is in x 1/2 is not a non-negative integer.

∴ 1 + √x + x are not a polynomial.

(ii) Here, the power of x in every term of the given expression is non-negative.

∴ 1 + x2 is a polynomial.

(iii) The power of x in the term 1/x, that is in, x–1 is a negative integer ∴ x + 1/x is not a polynomial.


Ans.

is a polynomial; since the degree of x in each term of the given expression is a non-negative integer. The degree of this polynomial is 2 since the maximum power of x in the given expression is 2.

is not a polynomial, since in 1/x2, that is in x–2, the power of x is a negative integer.

Short Answer Type questions: [3 Marks]

1. Check whether the following equations are quadratic or not :

(i) (x – 1)(x – 3) = (x + 5)(x – 1)

(ii) x2 + 2x – 6 √x + 5 = 0

(iii) x3 – 4x2 + 1 = (x – 2)3

Sol. (i) (x – 1) (x – 3) = (x + 5) (x – 1)

⇒ x2 – 4x + 3 = x2 + 4x – 5

⇒ –4x – 4x + 3 + 5 = 0

⇒ –8x + 8 = 0

Which is not of the form ax2 + bx + c = 0.

Thus, (x – 1) (x – 3) = (x + 5) (x – 1) is not a quadratic equation.

(ii) x2 + 2x – 6 √x + 5 = 0

The given equation contains a term involving √x, i.e., x 1/2.

Since, x2 + 2x – 6 √x + 5 is not a quadratic polynomial, therefore,

x2 + 2x – 6 5 x + = 0 is not a quadratic equation.

(iii) x3 – 4x2 + 1 = (x – 2)3

⇒ x3 – 4x2 + 1 = x3 – 8 – 6x(x – 2)

⇒ –4x2 + 1 = –8 – 6x2 + 12x

⇒ –4x2 + 6x2 – 12x + 1 + 8 = 0

⇒ 2x2 – 12x + 9 = 0, which is a quadratic equation as it is of the form

ax2 + bx + c = 0. Thus, the given equation is a quadratic equation.

2. Determine whether the given values are solutions (roots) of the equation or not.

(ii) a2x2 – 3abx + 2b2= 0, x = a/b, x=b/a

⇒    2 + 2 – 4 = 0

⇒ 4 – 4 = 0

⇒ 0 = 0, which is true.

x = √2 is a solution of the equation.

Now, putting x = −2 √2, we get

⇒   8 – 4 – 4 = 0

⇒ 0 = 0, which is true.

So, x = −2 √2 is a solution of the equation.

(ii) a2x2 – 3abx + 2b2 = 0

Putting x = a/b in the given equation, we get

⇒ a4 – 3a2b2 + 2b4 = 0, which is not true.

 x = a/b is not a solution to the given equation.

Again, by putting x = b/a in the given equation, we get

⇒  b2 – 3b2 + 2b2 = 0

⇒ –2b2 + 2b2 = 0

⇒ 0 = 0, which is true.

x = b/a is a solution of the given equation.

the following, find the value of k for which the given value is a solution of the given equation :

(i) x2 – x(a + b) + k = 0, x = a

(ii) kx2 + √2x – 4 = 0, x = 2

Sol. (i) Since x = a is a root of the given equation. Therefore, it satisfies the equation.

i.e., a2 – a (a + b) + k = 0

⇒ a2 – a2 – ab + k = 0

⇒ k = ab (ii) Since x = √2 is a root of the given equation. Therefore, it satisfies the equation.

⇒  b2 – 3b2 + 2b2 = 0

⇒ –2b2 + 2b2 = 0

⇒ 0 = 0, which is true.

x = b/a is a solution of the given equation.

3. In each of the following, find the value of k for which the given value is a solution of the given equation :

(i) x2 – x(a + b) + k = 0, x = a

(ii) kx2 + √2x – 4 = 0, x = 2

Sol. (i) Since x = a is a root of the given equation. Therefore, it satisfies the equation.

i.e., a2 – a (a + b) + k = 0

⇒ a2 – a2 – ab + k = 0

⇒ k = ab

(ii) Since x = √2 is a root of the given equation. Therefore, it satisfies the equation.

⇒ 2k + 2 – 4 = 0

⇒ 2k – 2 = 0 ⇒ k = 1

Long Answer Type questions: [4 Marks]

1. If x = – 2 and x = − 1/5 are solutions of the equation 5x2 + kx + λ = 0, find the values of k and λ.

Sol. Since x = –2 and x = − 1/5 are solutions of the equation, 5×2 + kx + λ = 0

∴ 5(–2)2 + k(–2) + λ = 0

⇒ 2k – λ = 20 …(i)

From (i), we get
λ = 2k − 20
Substituting this value of λ in (ii), we get

k − 5(2k − 20) = 1
⇒ k −10k +100 = 1
⇒ –9k = –99 ⇒ k = 11
∴ λ = 2k – 20 = 2(11) – 20 = 2
Hence, k = 11 and λ = 2

2. If x = 2/3 and x = –3 are the roots of the equation ax2 + 7x + b = 0, find the values of a and b. Sol. Since x = 2/3 is a root of ax2 + 7x + b = 0, we have,

⇒ 4a + 42 + 9b = 0

⇒ 4a + 9b + 42 = 0 …(i)

Again, x = –3 being a root of ax2 + 7x + b = 0, we have, a(–3)2 + 7(–3) + b = 0

⇒ 9a – 21 + b = 0

⇒ 9a + b – 21 = 0 …(ii)

Solving (i) and (ii), we get

a = 3 and b = – 6

3. If a and b are the roots of the quadratic equation x2 – p ( x + 1) – c = 0, then find the value of (a + 1) (b + 1).

Sol. We have x2 – p(x + 1) – c = 0

⇒ x2 – px – (p + c) = 0

⇒ Since a and b are roots of the equation

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